In Cantor's diagonal argument he assumes that the real numbers over a segment of the continuum [0.(0) to 0.(9)] can be put into a set and then assembled into a list in arbitrary order. One can easily observe that this works for pool balls: place the 15 balls into a bag, reach in and draw out one ball at a time, and line them up on the pool table in the order extracted. Cantor then proceeds to show that he can create a real number not in the list by:1. converting each number in the assumed complete list to binary.2. Taking the nth bit of the nth entry in the list and negating it.3. Taking the negated bits and stringing them together in order to form a constructed real number.

Whereas each bit in order in the constructed number differs from the corresponding bit in each real number in the same position in the list, the constructed number differs from all numbers in the list (i.e. the nth bit of the constructed number differs from the nth bit of the nth number in the list).

However, Cantor employed shallow thinking, and did not recognize the error in his argument. As he traverses the list, he creates intermediate numbers which are not in the currently traversed list.

The first bit of the first number in the list can be 0 or 1, so Cantor creates either 0.1(0) or 0.0(0)

The second bit of the second number in the list can be 0 or 1, so Cantor creates either 0.11(0), 0.10(0). 0.01(0) or 0.00(0)

The third bit of the third number in the list can be 0 or 1, so Cantor creates either 0.111(0), 0.110(0), 0.101(0), 0.100(0), 0.011(0), 0.010(0), 0.001(0) or 0.000(0)

And similarly for all other positions. Make note that the numbers created in sequence not in the list so far traversed are all rational fractions with finite significant bits and trailing zeros. Whereas there are only a finite number of rational fractions with finite significant bits (i.e. 2^n), all of these so far missing rational fractions must appear at later positions somewhere in the list of real numbers presumed to be complete; because the rational numbers are a subset of the real numbers, and the rational numbers with finite significant bits are a subset of the rational numbers in general.

For as long as Cantor is traversing the list of real numbers, creating rational fractions with finite significant bits not currently in the traversed list, the numbers created must eventually show up in the list. Whereas there are infinite real numbers in the segment 0.(0) to 0.(9), Cantor is endlessly traversing the list (i.e. there is no end to the list).

Although Cantor is only always approaching a fictitious end of the list, he assumes that he can get to the end (ignoring that it is fictitious); this is the fatal flaw in his argument. Were he able to do so he would have created an irrational number of infinite bits certifiably not in the list of reals. However, contradiction and paradox arises as to how it is possible to have an assumed complete list of reals, assembled from an assumed complete set, which is missing an element [1]. Answer: it is not possible. Whereas it is impossible for Cantor to get to the end of an endless list, the implied paradox is irrelevant because it is the consequence of an unobtainable situation: i.e. Cantor never does create the real number certifiably not in the list because it is impossible to do so by getting to the non-existent end of an endless list. The hypothesized real number certifiably not in the list does not exist: i.e. it cannot be created and is otherwise fanciful.

Q.E.D.

[1] If there is an element missing from the list then that element must either still be in the set, or missing from the assumed complete set from the start.

Edit added 10/1/2015. typos corrected once - 13 months after robert 46 started this preposterous thread! - because he refuses to even consider how he misrepresents Cantor's proof. Anybody reading this for the first time can simply look at the end of the thread, and see that this so.

Here is a simplified, but pedantically-described-because-robert's-obstinateness-requires-it, version of the algorithm known as Cantor Diagonalization:

Assume you can have infinite sets; that is, collections of things that have a beginning, but no end.

One infinite set is the set of all counting numbers, {1,2,3,...}. I'll call this set C. The capital-bold-italics notation means it is a set.

Another is the set of all even numbers E, or a collection of distinct, infinite binary strings S.

To enumerate an infinite set means to associate every counting number in C with a unique element of the set, for every member of the set.

That is, you have a function x(n) that describes every member of a set X.

For example, E can be enumerated by the function e(n)=2*n. This may seem like a contradiction, that a subset of C can be enumerated, but it is a necessary property of infinite sets.

Robert incorrectly calls it a list. A list is the special case of a finite enumeration.

Assume you have a set of infinite binary strings S that can be enumerated.

This means you have also defined a function b(n,m) that returns a single bit for every pair of integers n and m.

You can now define every bit of an infinite bit string d. For every integer n, The nth bit of d is d(n)=1-b(n,n).

The complete string d is not in S; because for any n, the string s(n) in S has its nth bit different than d's nth bit.

This proves that if you can enumerate a set of bit strings S, then there is at least one bit string that is not in S. It also proves the inverted converse: If every possible bit string is in S, then you cannot enumerate it.

That is the algorithm robert 46 misrepresents. In particular:

This algorithm does not describe an order, or even a process.

It defines every bit at once, just like f(n)=2*n describes every even number at once.

It also "uses" - whatever robert thinks that means - every string in S at once, just like f(n)=2*n "uses" every integer n at once.

So anytime robert argues about "the section of the list traversed so far," he is misrepresenting the algorithm. The whole set is used at the same time.

Anytime he talks about subsections, he is misrepresenting the algorithm, because there is no subsectioning involved.

Anytime he talks about a finite number of bits in d, he is misrepresenting the algorithm. The algorithm defines an infinite number of bits, all at the same time.

Anytime he talks about comparing the "length of an array" to its width, he is misrepresenting the algorithm. There is no array. The (infinite) number of elements in set S is the only thing remotely similar to what he calls the length, and the infinite length of the strings is the width. You can't compare infinities without first developing an arithmetic of infinities. Which is what Cantor's proof provides a basis for, and robert claims is invalid.

Anytime he accuses Cantor of assuming S includes every real number, or goes on about decimal points, he is misrepresenting the proof. Any string that can be in S can be interpreted as a real number between 0 and 1, and Cantor does not use a radix point. And Cantor never, never, NEVER assumes in step 5 above that every possible string is included. Suggesting otherwise, as robert does, shows that he does not understand how this algorithm is used in the proof, and that he refuses to try.

Anytime he talks about applying the proof to rational numbers instead of real numbers, he is misrepresenting the algorithm. What I omitted above is that any element in S can be shown to represent a real number, as can the missing element d. If you restrict S so that any element in S represents a rational number, you can't show that d is rational. So you can't show a rational number must be nissing.

There may be other fallacies that robert repeats that I'll add later. Because any objection he raises will be based on one.

+++++

Original post:

robert 46 wrote:In Cantor's diagonal argument he assumes that the real numbers over a segment of the continuum [0.(0) to 0.(9)]

Cantor's proof says that if distinct sets of infinite binary digits are placed in an order S{1},...,S{N}, that there is always another set you can call S{N+1} that is not in that list. The proof does not involve numbers, only sets of binary digits.

A consequence of this theorem is that fractional real numbers (both rational and irrational) between 0 and 1 are not countable, since each S{i}={B(1),B(2),...} can be used to express such a fraction F(n) by the summation F(n)=B(n,1)/2+B(n,2)/4+B(n,2)/8+...

However, Cantor employed shallow thinking, and did not recognize the error in his argument.

No, Robert 46 employs shallow thinking, by refusing to acknowledge how Cantor's proof is about sequences, not numbers.

And by refusing to acknowledge that, while Cantor's proof can be applied to values outside of the range [0,1], any such application is based on the linear transformation R(n,X,Y)=X*F(n)+Y, where X and Y are finite numbers. So it can't be applied to the set of "all integers," since (1) There is no (X,Y) that produces them, and (2) The infinite nature of the sets Cantor used require fractions (rational and irrational) to be considered.

As he traverses the list, he creates intermediate numbers which are not in the currently traversed list.

He does not create numbers, intermediate or otherwise. He is always dealing with a finite number of sets of infinitely many binary digits.

Make note that the numbers created in sequence not in the list so far traversed are all rational fractions with finite significant bits and trailing zeros.

No, the ones you considered are finite. Cantor placed no such limitation on his sequences - he only showed a finite number of digits in his examples, and "created" only the first N elements needed to distinct sequences.

For as long as Cantor is traversing the list of real numbers, creating rational fractions with finite significant bits not currently in the traversed list, the numbers created must eventually show up in the list.

He has no set "list" that he is "traversing." He uses examples to represent arbitrary lists. Robert should understand this concept, since he misuses it for probability puzzles.

Cantor does not "create rational fractions." You can create fractions from his sequences, and you can even limit them to the rationals if you want to, but Cantor does not. He creates the first N digits of an infinitely-long set of digits.

And his whole point was that there are always infinite sequences that do not appear in any finite list of N sequences. These are found by fixing the first N digits of those infinite sequences, but the (infinitely many) other digits are left arbitrary.

Although Cantor is only always approaching a fictitious end of the list,...

No, he really isn't:

"If s1, s2, … , sn, … is any enumeration of elements from T [the set of all infinite sequences of binary digits], then there is always an element s of T which corresponds to no sn in the enumeration."

this is the fatal flaw in his argument.

No, it is the flaw in your misunderstanding of the argument.

Whereas it is impossible for Cantor to get to the end of an endless list,

It is only you who says he tries.

the implied paradox is irrelevant because it is the consequence of an unobtainable situation:

The implied paradox is that, while there are an infinite number of integers you could use in an enumeration:

You can't place these sequences into a one-to-one correspondence with them.

You can't place these sequences into a finitely-many-to-one correspondence with them.

You can only place these sequences into an infinitely-many-to-one correspondence with them.

The paradox then becomes thinking of infinity as a number, since this shows that there are different "sizes" of infinities. The relationship sometimes used, 2^infinity, does not express a number, it expresses this abstract relationship.

Last edited by JeffJo on Fri Oct 02, 2015 6:44 am, edited 2 times in total.

robert 46 wrote:As usual, JeffJo is evading the issues with diversions. If he wishes to address my argument by quoting it statement-by-statement to identify any flaws in it in context, then I invite him to try.

As usual, robert evades the fact that I pointed out his errors, in context, by ignoring the error and making personal attacks. Makes one wonder what his definition of "evasion" is.

The question one must ask is, who is more believable? The mathematician who is considered the father of the modern concept of infinity, whom no educated mathematician has found fault with over 120 years; or an untrained crank? Of course, and also as usual, robert will come back with more insults about the profession of mathematician in general, but that will only reinforce my point.

I have presented a thesis. If JeffJo, or anyone, wants to challenge it, the ground rules are:1. Address my thesis per the context I have presented.2. For every statement I have made, either agree or disagree.3. If you disagree, provide an explanation for why you believe a statement is in error.4. Summarize why you believe my conclusion does not follow from my analysis.

After 5 1/2 years of studying JeffJo, in this matter I am not going to tolerate the trickery he characteristically uses, and be mired in banter such as JeffJo and Chuck have for weeks been engaged in in the Magazine Column topic Unequal Work: which is ludicrously "much ado about nothing".

If you want rules, how about "stop ignoring points you can't refute," "consider how you might be wrong before responding," and "if every expert in the field disagrees with you, you probably are wrong so stop insulting them for your fault."

But if you want a more concise criticism, you didn't say anything about Cantor's proof. You addressed an incorrect interpretation of it. If you want a better analysis of why you are wrong, address the actual proof.

Hint:

Construct two unit circles, C1 and C2, on the same page.

Let O1 and O2 be their centers.

Choose an arbitrary point P inside (or on) C2.

Draw line segment O1P.

Find the point Q on O1P such that len(O1Q)=1/len(O1P).

Repeat (this is a abstract iteration, not a real one, since it is endless) for every point inside of (or on) C2.

The result is a set of points {P} that constitute all of the points in C2, and a set {Q} that constitute a subset of the points in C1. Thus "proving" that there are more points in C2 than in C1.

But the reverse procedure, mapping all possible Q's to a P, "proves" that there are more points in C1 than in C2.

This is a contradiction. There must be an invalid assumption somewhere. What is it? How is this different than what you did (answer: it isn't). And how are both different from what Cantor did (for the answer, see my first reply).

...and a list of JeffJo's posts in the other hand, and match the left with the right.

... and find none. Robert's, however, have many matches. The first, "Ad hominem attacks" is a big one - just look for any time he disagrees with established experts. The second, "Affirming The Consequent" is the basis for his alternate theory of probability.

Astute readers should be able to deduce from JeffJo's evasions and diversions that:1. He well-understands that he cannot refute my thesis in close combatSo:2. He resorts to sniping away from afarBut:3. He is clueless about the fundamental issue which the thesis raises, and his potshots can't hit the broad side of a barn.

robert 46 wrote:Astute readers should be able to deduce from JeffJo's evasions and diversions that:1. He well-understands that he cannot refute my thesis in close combatSo:2. He resorts to sniping away from afarBut:3. He is clueless about the fundamental issue which the thesis raises, and his potshots can't hit the broad side of a barn.

Actually, what is easily deduced is:

Robert didn't address Cantor's proof, he addressed a misinterpretation of it.

Since Robert can't refute fact A, which was well expressed, he evades the issue by:

Taking potshots at me, like this last one.

Insisting on rules designed to evade the issue (there was no valid context to address, there was no point to agreeing or disagreeing with his statements that didn't address Cantor's proof, etc.).

I addressed the diagonal argument in the context of real numbers: there is no meaningful-to-the-argument difference between a sequence of bits, and a sequence of bits given significance by positional notation.

JeffJo says that Cantor is dealing with a finite list. Perhaps JeffJo has forgotten the meaning of ellipsis dots. He tries to project his misunderstanding of the diagonal argument onto me. Amusing.

JeffJo (re)introduces the two circles example, but fails to explain how it is relevant to the diagonal argument. It is a red-herring diversion.

What JeffJo says which is most informative about himself occurs here, however:

robert 46 wrote: ...the ground rules are...

JeffJo wrote: Disagree.

JeffJo doesn't acknowledge allegiance to rules. That is why his posts are peppered with fallacies of debate.

***** 2014-09-11

JeffJo wrote:He [Cantor] is always dealing with a finite number of sets of infinitely many binary digits.

...his [Cantor's] whole point was that there are always infinite sequences that do not appear in any finite list of N sequences.

No one is alarmed by infinite sequences being missing from a finite set. That is not Cantor's argument.

Last edited by robert 46 on Thu Sep 11, 2014 9:46 am, edited 2 times in total.

robert 46 wrote:I addressed the diagonal argument in the context of real numbers:

No, as I've explained several times, you addressed an incomplete misinterpretation of it. The theorem is:

For every set S, the power set of S (i.e., the set of all subsets of S) has a larger cardinality than S itself.

Binary sequences and real numbers are ways of visualizing it, but are not the theorem itself.

JeffJo says that Cantor is dealing with a finite list.

No, Jeffjo said nothing about whether the list was finite or infinite. The general theorem applies to either. The visualization with reals requires that the elements of the list be infinite sequences.

Perhaps (well, it's been demonstrated many times) robert can't read for comprehension, when he wants to find something other than what was said.

JeffJo (re)introduces the two circles example, but fails to explain how it is relevant to the diagonal argument.

If you answer the questions, you will see.

JeffJo doesn't acknowledge allegiance to rules.

No, I don't agree that you get to set them in such a narrow fashion in a discussion forum. Especially since your so-called rules prevented the valid counter-arguments I presented. Which is what they were designed, by you, to do.

A consequence of this theorem is that fractional real numbers (both rational and irrational) between 0 and 1 are not countable, since each S{i}={B(1),B(2),...} can be used to express such a fraction F(n) by the summation F(n)=B(n,1)/2+B(n,2)/4+B(n,2)/8+...

JeffJo, all rational numbers that are in fractional form can be put into a one-to-onecorrespondence with the set of positive integers (the counting numbers). The fractionalrational numbers are countable.